[13] Roman to Integer

© Jarvus Chen / www.jarvus.net

Description

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

Required knowledge

Roman numerals conversion table

• I = 1
• V = 5
• X = 10
• L = 50
• C = 100
• D = 500
• M = 1000

My solution ( Beat 96.96% )

addis the move distance for the next position in string.

int romanToInt(char* s) {

    char *data = "IVXLCDM00";
    int value[9]= {1,5,10,50,100,500,1000,0,0};    

    int ans = 0;
    int i = 0;
    int add = 1;
    int size = strlen(s);

    while ( i < size ){
        add = 1;
        for ( int j = 0 ; j < 7 ; j++){
            if ( s[i] == data[j] ){
                if ( i != size ){
                    if ( s[i+1] == data[j+1] ){
                        ans = ans + value[j+1] - value[j];
                        add = add + 1;
                    }else if( s[i+1] == data[j+2] ){
                        ans = ans + value[j+2] - value[j];
                        add = add + 1;
                    }else{
                        ans = ans + value[j];
                    }              
                }
            }  
        }
        i = i + add;
    }
    return ans;
}

Others solution

https://discuss.leetcode.com/topic/71045/a-c-solution-easy-to-understand

Use weight as the index for all ACSII.

int romanToInt(char* s) {
    int weight[256] = {0}; // store weights of I,V,X,L,C,D,M
    weight['I'] = 1;
    weight['V'] = 5;
    weight['X'] = 10;
    weight['L'] = 50;
    weight['C'] = 100;
    weight['D'] = 500;
    weight['M'] = 1000;
    int weight_pre = 0;
    int i = strlen(s) - 1;
    int num = 0;
    while(i >= 0) {
        if(weight[s[i]] >= weight_pre)
            num += weight[s[i]];
        else
            num -= weight[s[i]];
        weight_pre = weight[s[i--]];
    }
    return num;
}